Cho 1/x+1/y+1/z=0.Tính S=xy/z^2+yz/x^2+xz/y^2

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{1}{x} +\frac{1}{y} +\frac{1}{z} =0\\
\Rightarrow \frac{1}{x} +\frac{1}{y} =\frac{-1}{z}\\
\Rightarrow \left(\frac{1}{x} +\frac{1}{y}\right)^{3} =\frac{-1}{z^{3}}\\
\Rightarrow \frac{1}{x^{3}} +3.\frac{1}{x^{2}} .\frac{1}{y} +3.\frac{1}{x} .\frac{1}{y^{2}} +\frac{1}{y^{3}} =\frac{-1}{z^{3}}\\
\Rightarrow \frac{1}{x^{3}} +\frac{1}{y^{3}} +\frac{1}{z^{3}} =\frac{-3}{xy}\left(\frac{1}{x} +\frac{1}{y}\right)\\
\Rightarrow \frac{1}{x^{3}} +\frac{1}{y^{3}} +\frac{1}{z^{3}} =\frac{-3}{xy} .\frac{-1}{z}\\
\Rightarrow \left(\frac{1}{x^{3}} +\frac{1}{y^{3}} +\frac{1}{z^{3}}\right) xyz=3\\
\Rightarrow \frac{yz}{x^{2}} +\frac{xz}{y^{2}} +\frac{xy}{z^{2}} =3\\
\Rightarrow S=3
\end{array}$